TThe Ronkin number of an exponential sum.
James Silipo ∗ October 29, 2010
Abstract
We give an intrinsic estimate of the number of connected compo-nents of the complementary set to the amoeba of an exponential sumwith real spectrum improving the result of Forsberg, Passare and Tsikhin the polynomial case and that of Ronkin in the exponential one.
Keywords: amoeba, exponential sum, Ronkin function, Jessen func-tion, Ronkin number.
Let Ω ⊆ R n be a non empty open convex subset and let T Ω = Ω + i R n ⊆ C n be the (vertical) tube domain on the base Ω. Let also Re : T Ω → Ω be theprojection of T Ω onto its base. The amoeba F Y of a closed analytic subset Y of T Ω is the topological closure in Ω of Re Y , i.e. F Y = Re Y . (1)This notion of amoeba was originally proposed by Favorov [2] for zerosets of holomorphic almost periodic functions defined on tube domains.Recall that a holomorphic function f defined on a tube domain T Ω ⊆ C n is said to be almost periodic if, for every D (cid:98) Ω, f is the uniform limit on T D = D + i R n of a sequence of exponential sums, i.e. C -linear combinationsof exponentials e (cid:104) z,λ (cid:105) , with z ∈ C n , λ ∈ R n and (cid:104) z, λ (cid:105) = z λ + . . . + z n λ n .For any non empty open and convex subset Ω ⊆ R n , the class of holomorphic ∗ The author was partially supported by the grant n. A1UNICAL053 - POR Calabria2000-2006 Misura 3.7 Azione B. a r X i v : . [ m a t h . C V ] S e p lmost periodic functions on T Ω will be noted by HAP ( T Ω ), the subclass ofexponential sums will be noted by Exp( T Ω ).Any f ∈ HAP ( T Ω ) has a well defined Bohr transform a ( f, · ) : R n → C given, for every λ ∈ R n , by a ( f, λ ) = lim s → + ∞ s ) n (cid:90) | y (cid:96) |
Let f be an exponential sum on C n , then F f = R n ∩ (cid:91) χ ∈ Ch Ξ f V ( f χ ) = (cid:91) χ ∈ Ch Ξ f Re V ( f χ ) , (19) in particular F f = F f χ , for every χ ∈ Ch Ξ f . Another easy though useful fact about the shape of an exponential amoebais given by the following proposition.
Proposition 2.1
Let f be an exponential sum defined on C n , then F f = F f + (lin Γ f ) ⊥ , (20) where (lin Γ f ) ⊥ is the orthogonal complement of the linear subspace spannedby Γ f in R n endowed with the standard scalar product. Proof.
Let χ ∈ Ch Ξ f and consider the zero set V ( f χ ) of f χ in C n . If π f isthe linear projection of C n onto the complex linear subspace (lin Γ f + i lin Γ f ),then, the expression of f χ implies that, z ∈ V ( f χ ) ⇐⇒ π f ( z ) + (lin Γ f + i lin Γ f ) ⊥ C ⊆ V ( f χ ) , (21)where ⊥ C stands for orthogonality in C n with its standard hermitian prod-uct. Taking real parts yields z ∈ V ( f χ ) = ⇒ Re π f ( z ) + (lin Γ f ) ⊥ ⊆ Re V ( f χ ) , (22)where the equality Re (lin Γ f + i lin Γ f ) ⊥ C = (lin Γ f ) ⊥ follows by the obviousinclusion Γ f ⊂ R n . As a consequence (cid:91) z ∈ V ( f χ ) Re π f ( z ) + (lin Γ f ) ⊥ ⊆ Re V ( f χ ) , (23)6nd taking the union on Ch Ξ f implies that (cid:91) χ ∈ Ch Ξ f (cid:91) z ∈ V ( f χ ) Re π f ( z ) + (lin Γ f ) ⊥ = (lin Γ f ) ⊥ + (cid:91) χ ∈ Ch Ξ f (cid:91) z ∈ V ( f χ ) Re π f ( z )= (lin Γ f ) ⊥ + (cid:91) χ ∈ Ch Ξ f Re V ( f χ )= (lin Γ f ) ⊥ + F f , so F f + (lin Γ f ) ⊥ has to be included in F f thus proving what claimed.The idea behind our counting technique is to linearly embed an expo-nential amoeba into a naturally associated polynomial one so as to obtainan intrinsic estimate of the Ronkin number. We start by constructing theassociated polynomial amoeba.Let f ∈ Exp( C n ) and let γ : Ξ f → Z rank Ξ f be a group isomorphism. Inthe vector space Z rank Ξ f ⊗ Z R consider the convex hull conv ( γ (Sp f )) of theimage of Sp f via γ and the setΛ f = γ − (conv ( γ (Sp f )) ∩ γ (Ξ f )) . (24)Since γ (Ξ f ) = Z rank Ξ f , the set Λ f is the subset of Ξ f consisting of the inverseimages via γ of the lattice points belonging to the polytope conv ( γ (Sp f )).Evidently, among these lattice points there are the elements of Sp f , so thatSp f ⊆ Λ f , but in fact one can say more. Lemma 2.1
Let f ∈ Exp( C n ) with rank Ξ f = r and let γ : Ξ f → Z r be agroup isomorphism. Then Λ f ⊂ Γ f . Proof.
Let µ ∈ Λ f , then there exists a t ∈ [0 , Sp f summing up to 1 suchthat γ ( µ ) = (cid:88) λ ∈ Sp f t λ γ ( λ ) ∈ conv ( γ (Sp f )) ∩ Z r . (25)For every 1 (cid:54) (cid:96) (cid:54) r , let ω (cid:96) the inverse image via γ of the (cid:96) -th element e (cid:96) ofthe canonical basis of Z r , then, for every λ ∈ Sp f , λ = r (cid:88) (cid:96) =1 λ (cid:96) ω (cid:96) , γ ( λ ) = r (cid:88) (cid:96) =1 λ (cid:96) e (cid:96) (26)7nd γ ( µ ) = r (cid:88) (cid:96) =1 (cid:18) (cid:88) λ ∈ Sp f t λ λ (cid:96) (cid:19) e (cid:96) ∈ Z r , (27)so that µ = γ − r (cid:88) (cid:96) =1 (cid:18) (cid:88) λ ∈ Sp f t λ λ (cid:96) (cid:19) e (cid:96) (28)= r (cid:88) (cid:96) =1 (cid:18) (cid:88) λ ∈ Sp f t λ λ (cid:96) (cid:19) ω (cid:96) (29)= (cid:88) λ ∈ Sp f t λ (cid:18) r (cid:88) (cid:96) =1 λ (cid:96) ω (cid:96) (cid:19) (30)= (cid:88) λ ∈ Sp f t λ λ ∈ Γ f . (31)By the arbitrary choice of µ in Λ f we obtain Λ f ⊂ Γ f .The following lemma implies that Λ f does not depend on the isomor-phism used in its definition. Lemma 2.2
Let f be an exponential sum and let γ and γ be two differentisomorphisms of the group Ξ f on Z rank Ξ f . Then the polytopes conv ( γ (Sp f )) and conv ( γ (Sp f )) are combinatorially isomorphic. Proof.
Let r = rank Ξ f . The authomorphism of Z r given by γ ◦ γ − hasan obvious continuation to an R -linear automorphism of R r . The two latticepolytopes conv ( γ (Sp f )) and conv ( γ (Sp f )) correspond to each other inthis R -linear authomorphism, hence they are combinatorially isomorphic.Since a combinatorial isomorphism of lattice polytopes preserves latticepoints in the polytopes, the set Λ f proves to depend only on f . As shownin the sequel of the article, the interest in the set Λ f is due to the role of“order set” it will play for R n \ F f .Now, if r = rank Ξ f , fix an isomorphism γ : Ξ f → Z r and for every1 (cid:54) j (cid:54) r , let ω j be the inverse image via γ of the j -th element of the8anonical basis of Z r . This basis and the isomorphism γ will be referred toas associated to each other. Then consider the Laurent polynomial P ( ζ ) = (cid:88) λ ∈ Sp f a ( f, λ ) ζ γ ( λ ) = (cid:88) k ∈ γ (Sp f ) a ( f, γ − ( k )) ζ k . . . ζ k r r , (32)and observe that for any character χ ∈ Ch R n we have f χ ( x ) = P (cid:0) e (cid:104) x, ω (cid:105) + i Arg χ ( ω ) , . . . , e (cid:104) x, ω r (cid:105) + i Arg χ ( ω r ) (cid:1) , (33)for every x ∈ R n . Consider also the linear mapping L : R n → R r given by L ( x ) = ( (cid:104) x, ω (cid:105) , . . . , (cid:104) x, ω r (cid:105) ) , (34)and notice that the kernel of L equals the orthogonal complement (withrespect to the standard scalar product of R n ) of the linear subspace lin Γ f spanned by Γ f . It follows thatdim( L ( R n )) = n − dim ker L = dim lin Γ f , (35)in particular, the mapping L is injective if and only if the linear sub-space lin Γ f is full-dimensional.The following lemma gives some additional information about this con-struction. Lemma 2.3
Let f ∈ Exp( C n ) and γ : Ξ f → Z r be an isomorphism.Let also P be the corresponding Laurent polynomial, Σ P the normal fanof Γ P , h Γ P the support function of the polytope Γ P and, for every face ∆ of Γ P , let K ∆ , Γ P be the corresponding dual cone. Then1. L ( R n ) \ { } does not intersect the cones of Σ P corresponding to thepositive dimensional faces of Γ P ,2. h Γ P ( L ( x )) = h Γ f ( x ) , (36) for every x ∈ R n . In particular, for every λ ∈ Vert(Γ f ) , L ( K λ, Γ f ) ⊆ K γ ( λ ) , Γ P . (37)9 . L ( R n ) ⊂ { } ∪ (cid:91) λ ∈ Vert(Γ f ) int K γ ( λ ) , Γ P . (38) Proof.
1. Let { ω , . . . , ω r } be the basis of Ξ f associated to γ and suppose,by contradiction, there is an x ∈ R n \ ker L such that L ( x ) belongs to acone of Σ P corresponding to a positive dimensional face ∆ of Γ P . Since ∆is positive dimensional, it admits two distinct lattice points u and v , so thevectors t = u − v and L ( x ) are orthogonal to each other. This means that0 = (cid:104) t, L ( x ) (cid:105) R r = r (cid:88) (cid:96) =1 t (cid:96) (cid:104) x, ω (cid:96) (cid:105) R n = r (cid:88) (cid:96) =1 (cid:104) x, t (cid:96) ω (cid:96) (cid:105) R n = (cid:104) x, r (cid:88) (cid:96) =1 t (cid:96) ω (cid:96) (cid:105) R n . (39)Since ω , . . . , ω r are Z -linear independent and t (cid:54) = 0, the preceding equalitiesmake x to belong to (lin Γ f ) ⊥ , which equals the kernel of L . The contradic-tion implies the statement.2. Let x ∈ R n be fixed, then by definition of support function h Γ P ( L ( x )) = sup v ∈ Γ P (cid:104) L ( x ) , v (cid:105) R r = sup v ∈ Vert(Γ P ) (cid:104) L ( x ) , v (cid:105) R r . (40)However Vert(Γ P ) ⊆ γ (Sp f ), so h Γ P ( L ( x )) = sup λ ∈ Sp f (cid:104) L ( x ) , γ ( λ ) (cid:105) R r . (41)Any λ ∈ Sp f has an expression of the form λ = r (cid:88) (cid:96) =1 k λ,(cid:96) ω (cid:96) , (42)for a uniquely determined sequence k λ, , . . . , k λ,r of integers, then h Γ P ( L ( x )) = sup λ ∈ Sp f r (cid:88) (cid:96) =1 (cid:104) x, ω (cid:96) (cid:105) R n k λ,(cid:96) = sup λ ∈ Sp f (cid:104) x, r (cid:88) (cid:96) =1 k λ,(cid:96) ω (cid:96) (cid:105) R n = sup λ ∈ Sp f (cid:104) x, λ (cid:105) R n , (43)i.e. h Γ P ( L ( x )) = h Γ f ( x ).In particular, if λ ∈ Γ f and x ∈ K λ, Γ f , then (cid:104) L ( x ) , γ ( λ ) (cid:105) R r = (cid:104) x, λ (cid:105) R n = h Γ f ( x ) = h Γ P ( L ( x )) , (44)10.e. L ( K λ, Γ f ) ⊆ K γ ( λ ) , Γ P .3. The whole R n is equal to the union of the closures of the cones whichare dual to the vertices of Γ f and each of these cones is mapped by L in theinterior of the dual cone associated to the corresponding vertex in Γ P . Theorem 2.2
Let f ∈ Exp( C n ) and let γ : Ξ f → Z r be an isomorphismwith the associated basis { ω , . . . , ω r } of Ξ f . If P is the corresponding Lau-rent polynomial and L the corresponding linear mapping, then1. the following two equalities hold true L ( F f ) = L ( R n ) ∩ A P and L ( R n \ F f ) = L ( R n ) \ A P , (45)
2. the Ronkin number ρ ( f ) equals the number of connected componentsof L ( R n ) \ A P ,3. the following estimate holds true f ) (cid:54) ρ ( f ) (cid:54) f . (46) Proof.
1. Let us start with the first equality. If x ∈ F f , then, by equal-ity (19), there exists a character χ ∈ Ch Ξ f such that f χ ( x ) = 0. By (33)it follows that L ( x ) ∈ A P . Conversely, if x ∈ R n is such that L ( x ) belongsto A P , then there is a zero ζ of P for which Log ( ζ ) = L ( x ), i.e. ln | ζ (cid:96) | = (cid:104) x, ω (cid:96) (cid:105) , for 1 (cid:54) (cid:96) (cid:54) r . Let χ ∈ Ch Ξ f be the uniquely defined character suchthat χ ( ω (cid:96) ) = e i Im ζ (cid:96) , for 1 (cid:54) (cid:96) (cid:54) r , then0 = P ( ζ ) = f χ ( x ) , (47)or equivalently, x ∈ F f .As for the second equality, if x does not belong to F f , then f χ ( x ) (cid:54) = 0,for any character. Suppose, by contradiction, that L ( x ) belongs to A P .Then L ( x ) = Log ( ζ ), for some zero ζ of P . Let ϑ ∈ [0 , π ) r be the r -tupleof principal arguments of ζ and let also χ ∈ Ch Ξ f be the correspondingcharacter. By virtue of (33), f χ ( x ) = 0 and this contradicts the choice of x ,thus L ( R n \ F f ) ⊆ L ( R n ) \ A P . Finally, let x ∈ R n such that L ( x ) / ∈ A P ,then for every ϑ ∈ [0 , π ) r , P (cid:0) e (cid:104) x, ω (cid:105) + iϑ , . . . , e (cid:104) x, ω r (cid:105) + iϑ r (cid:1) (cid:54) = 0 , (48)11quivalently, (again by (33)), f χ ( x ) (cid:54) = 0 for any χ ∈ Ch Ξ f , whence x / ∈ F f .2. The connected components of L ( R n ) \ A P are convex. Let m be thenumber of such components. Since L is continuous, by Lemma 2.2, we knowthat ρ ( f ) (cid:62) m . Suppose, by contradiction, ρ ( f ) > m . Then we may findtwo points x and x belonging to distinct components of R n \ F f which aremapped by L to a same component Y of L ( R n ) \ A P . By virtue of Propo-sition 2.1, the projections x (cid:48) and x (cid:48) on lin Γ f of x and x respectively arestill distinct and belonging to F f . Now, ker L = (lin Γ f ) ⊥ , so the restric-tion of L to F f ∩ lin Γ f realizes a linear homeomorphism onto L ( R n ) \ A P .If α ⊂ Y is a line segment joining L ( x ) and L ( x ), then L − ( α ) ∩ lin Γ f isa line segment joining x (cid:48) and x (cid:48) , thus Proposition 2.1 makes x and x tobelong to the same component of R n \ F f . The contradiction implies thecorollary.3. The results of [3] imply that the number of components of R r \ A P cannot exceed P ∩ Z r ). Since the linear subspace L ( R n ) may intersect allthe components of R r \ A P , by Lemma 2.2 we get the desired estimate fromabove. The lower bound is easily found by the usual geometric series trick.It should perhaps be mentioned that the estimate (46) completely ne-glects the values of the coefficients of f except to those which correspond tothe vertices of Γ f , on which the only requirement is to be non-zero. A betterestimate would involve in a more substantial way the values of the Fourier co-efficients of f . Nevertheless, we notice that when Sp f ⊆ Z n , the present esti-mate improves the well known result of Forsberg, Passare and Tsikh [3] sincethe upper bound f does not exceed f ∩ Z n ). As an example, considerthe exponential sum f ( z ) = 2 + e z + e z + e z +4 z ∈ Exp( C ). Of courseΞ f = (2 Z ) , so if γ : Ξ f → Z is the isomorphism mapping (2 ,
0) to (1 , ,
2) to (0 , f ) = 5, whereas f ∩ Z ) = 11. A Laurent polynomial is called maximally sparse if all the points in thesupport of summation of the polynomial are vertices of its Newton polytope.Likewise, an exponential sum is called maximally sparse if all the pointsin its spectrum are vertices of its Newton polytope.12 group isomorphism γ : Ξ f → Z rank Ξ f cannot generally admit an R -linear continuation to R n , nevertheless, though such an isomorphism cannotpreserve the convex structure of Γ f , something still survives. The followingproposition is a key ingredient in order to relate the notion of maximalsparseness with the preceding construction. Proposition 3.1
Let f ∈ Exp( C n ) and let { ω , . . . , ω r } be the basis of Ξ f associated to some isomorphism γ : Ξ f → Z r together with the correspondingLaurent polynomial P . Then γ (Vert(Γ f )) ⊆ Vert(Γ P ) . (49) In particular, if f is maximally sparse, then the corresponding Laurent poly-nomial P is maximally sparse too. Proof.
Suppose, by contradiction, there is a λ ∗ ∈ Vert(Γ f ) such that γ ( λ ∗ )is not a vertex of Γ P . Since Γ P = conv ( γ (Sp f )), it follows that γ ( λ ∗ ) = (cid:88) λ ∈ Sp f \{ λ ∗ } t λ γ ( λ ) , (50)for some family { t λ } λ ∈ Sp f \{ λ ∗ } of non negative real numbers summing upto 1. For every λ ∈ Sp f there is a unique sequence k λ, , . . . , k λ,r of integerssuch that λ = r (cid:88) j =1 k λ,j ω j , (51)so γ ( λ ∗ ) = r (cid:88) j =1 k λ ∗ ,j γ ( ω j )and also γ ( λ ∗ ) = (cid:88) λ ∈ Sp f \{ λ ∗ } t λ r (cid:88) j =1 k λ,j γ ( ω j ) = r (cid:88) j =1 (cid:88) λ ∈ Sp f \{ λ ∗ } t λ k λ,j γ ( ω j ) . As { γ ( ω ) , . . . , γ ( ω r ) } is a basis of the vector space Z r ⊗ R , the two lastexpressions of γ ( λ ∗ ) yield the equality k λ ∗ ,j = (cid:88) λ ∈ Sp f \{ λ ∗ } t λ k λ,j , (52)13or every 1 (cid:54) j (cid:54) r . In each of these equalities the first member is an integerand so the second member has to be an integer too, it follows that r (cid:88) j =1 k λ ∗ ,j ω j = r (cid:88) j =1 (cid:88) λ ∈ Sp f \{ λ ∗ } t λ k λ,j ω j , or equivalently λ ∗ = r (cid:88) j =1 (cid:88) λ ∈ Sp f \{ λ ∗ } t λ k λ,j ω j = (cid:88) λ ∈ Sp f \{ λ ∗ } t λ r (cid:88) j =1 k λ,j ω j = (cid:88) λ ∈ Sp f \{ λ ∗ } t λ λ , (53)i.e. λ ∗ is not a vertex of Γ f . The contradiction implies the first statement.To prove the second statement, notice that for any f , Vert(Γ P ) ⊆ γ (Sp f ). If f is maximally sparse, then Sp f = Vert(Γ f ) and γ (Sp f ) = γ (Vert(Γ f )) = Vert(Γ P ) . (54)As γ (Sp f ) is precisely the support of summation of P the conclusion is thatall the elements in the support of summation of P are vertices of Γ P , i.e. P is maximally sparse.Observe that the second statement in Proposition 3.1 may not be re-versed. In fact the exponential sum f = 1 + e z + e √ z is not maximallysparse, whereas P = 1 + ζ + ζ is such.A polynomial (resp. exponential) amoeba is said to be solid if its com-plementary set has the minimal number of connected components.Equivalently a polynomial (resp. exponential) amoeba is solid if thenumber of connected components of its complementary set equals the num-ber of vertices in the Newton polytope of a Laurent polynomial (resp. ex-ponential sum) which defines the amoeba.Passare and Rullg˚ard [6] suggested the following conjecture. Conjecture 3.1 ([6])
A maximally sparse Laurent polynomial has a solidamoeba.
The conjecture is true if the Newton polytope of the given maximallysparse Laurent polynomial is reduced to a line segment, however it is notknown if the conjecture is true in the general case. Nisse [5] has recently14roposed a solution in the affirmative, but his proof seems to need someclarification. We notice here that such a solution would imply an exponentialcounterpart.
Proposition 3.2
If Conjecture 3.1 is true, a maximally sparse exponentialsum has a solid amoeba.
Proof.
Let f be a maximally sparse exponential sum on C n . With the samenotation of Proposition 3.1, the polynomial P is maximally sparse too soby Conjecture 3.1 the amoeba A P is solid. The estimate (46) for maximallysparse f and P becomes f ) (cid:54) ρ ( f ) (cid:54) P ) , (55)but the proof of Proposition 3.1 shows that the bounds in the above estimatecoincide so F f is solid. Let f ∈ Exp( C n ) and let E be a component of R n \ F f . According toFavorov [2], the order of E is the value ord ( E ) taken by the gradient of theJessen function (7) of f at any point of the domain E , i.e.ord ( E ) = grad J f ( x ) , (56)for any x ∈ E . Observe that the order of a complement component does notdepend merely on the amoeba. In fact, for every λ ∈ R n , the exponentialsum g ( z ) = e (cid:104) z,λ (cid:105) f ( z ) has the same amoeba as f but, for every complementcomponent E and any x ∈ E , one hasgrad J g ( x ) = λ + grad J f ( x ) . (57)In the almost periodic literature the order of a component E ⊂ R n \F f isknown as the mean motion of f in the domain E and, as shown in Ronkin [8],it is well defined. Here we propose a different approach to this order theory.Consider the Pontriagin group Ch Ξ f = Hom Z (Ξ f , S ). If rank Ξ f = r ,then Ch Ξ f is an r -dimensional real compact torus. Define the Ronkinfunction of f as the function N f : R n −→ R given, for any x ∈ R n , by N f ( x ) = (cid:90) Ch Ξ f ln | f χ ( x ) | dϑ ( χ ) , (58)15here ϑ ( χ ) is the translation invariant probability Haar measure on Ch Ξ f .The measure ϑ can be computed by choosing a basis { ω , . . . , ω r } of Ξ f andintegrating the differential form1(2 π ) r d Arg χ ( ω ) ∧ . . . ∧ d Arg χ ( ω r ) . (59)In order to realize that this definition is a good one, it is enough toconsider the Laurent polynomial P corresponding to a chosen basis of Ξ f and remark that, for any x ∈ R n , N f ( x ) = (cid:90) Ch Ξ f ln | P (cid:0) e (cid:104) x,ω (cid:105) + i Arg χ ( ω ) , . . . , e (cid:104) x,ω r (cid:105) + i Arg χ ( ω r ) (cid:1) | dϑ ( χ )= (cid:90) Log − ( L ( x )) ln | P ( ζ ) | dη r ( ζ ) , i.e. N f ( x ) = N P ( L ( x )) . (60)Thus the Ronkin function of f is the restriction of the Ronkin function of P to the subspace L ( R n ). As such N f proves to be well defined and convex.The function N f was already studied by Ronkin in [8] where he provedthe following theorem. Theorem 3.1 ([8], Theorem 6)
Let f be an exponential sum, then J f = N f . (61)Using the function N f and the equality (60) it is quite easy to prove thefollowing result. Lemma 3.1
Let f ∈ Exp( C n ) , then the Ronkin function N f is piecewiselinear on R n \ F f , its gradient mapping grad N f realizes an injection of theset of components of R n \ F f into Λ f . Proof.
Let { ω , . . . , ω r } be a basis of Ξ f and let γ be the associated iso-morphism of Ξ f on Z r . If P is the corresponding Laurent polynomial, thenfor every 1 (cid:54) j (cid:54) n and any x ∈ R n \ F f , ∂ N f ∂x j ( x ) = r (cid:88) (cid:96) =1 ∂N P ∂ζ (cid:96) (cid:0) L ( x ) (cid:1) ∂ ( L ) (cid:96) ∂x j ( x ) = r (cid:88) (cid:96) =1 ∂N P ∂ζ (cid:96) (cid:0) L ( x ) (cid:1) ω (cid:96),j , (62)16.e. grad x N f ( x ) = r (cid:88) (cid:96) =1 ∂N P ∂ζ (cid:96) ( L ( x )) ω (cid:96) = γ − (grad ζ N P ( L ( x )) . (63)Now, N P is piecewise linear on R r \ A P , its gradient mapping is constanton each component of R r \ A P and it maps injectively the set of thesecomponents on a finite subset of Γ P ∩ Z r . Consequently, N f is piecewiselinear on R n \F f , its gradient mapping is constant on each component of R n \F f and, as the ω , . . . , ω r are Z -linearly independent, it maps injectively theset of such components on a finite subset of R n . In order to identify thissubset observe that equation (63) impliesgrad x N f ( x ) ∈ γ − (grad ζ N P ( L ( R n ) \ A P )) ⊆ γ − (Γ P ∩ Z r ) = Λ f , (64)for any x ∈ R n \ F f .With respect to Favorov’s result, Lemma 3.1 actually adds that, for anyexponential sum f , the gradient of J f = N f maps the amoeba comple-ment R n \ F f in the Newton polytope Γ f , thus completing the analogy withthe polynomial case.It should also be noticed that, for a given f ∈ Exp( C n ), if r = rank Ξ f and γ : Ξ f → Z r is a fixed isomorphism then, up to a multiplication bya suitable exponential monomial, the corresponding Laurent polynomial P is an ordinary polynomial and the estimate ρ ( f ) (cid:54) υ ( f, γ ) provided byRonkin can be found by bounding the number of lattice points (with positivecoordinates) belonging to the ball about the origin of R r with radius equalto max x ∈ R r \A P grad N P ( x ) . (65)As the Newton polytope Γ P is properly contained in that ball, Lemma 3.1implies that f < υ ( f, γ ) . (66)It is also worth noting that, unlike υ ( f, γ ), the bound f ) is intrinsicsince it does not depend on the isomorphism used to compute it.17 Some examples
The following examples show the preceding constructions at work. For thesake of simplicity we just consider the case r = 2 and 1 (cid:54) n (cid:54) r . Thefigures placed at the end show, for each example, the Newton polytope Γ P ,the points in the subset γ (Λ f ) ⊂ Γ P and the amoeba A P cut by the linearsubspace L ( R n ). Example 4.1
Let f ∈ Exp( C ) be given by f ( z ) = 1 + 3 e √ z + e √ z . Thegroup Ξ f has rank r = 2, the numbers √ √ γ denotes the associated isomorphism of Ξ f on Z one gets - P ( ζ , ζ ) = 1 + 3 ζ + ζ , - < υ ( f, γ ) = 2 − π (cid:0) √ <
24 , - f = 3 . Γ P Figure 1: ρ ( f ) = f = 3. Example 4.2
Let f ∈ Exp( C ) be given by f ( z ) = 1 + e √ z + e πz − e ( √ π ) z . The group Ξ f has rank r = 2, the numbers 3 √ √ π generate it freely and if γ denotes the associated isomorphism of Ξ f on Z one obtains - P ( ζ , ζ ) = 1 + ζ + ζ − ζ − ζ , - < υ ( f, γ ) = 2 − π (cid:0) √ <
142 ,18 f = 4 . Γ P Figure 2: 3 = ρ ( f ) (cid:54) f = 4. Example 4.3
Let f ∈ Exp( C ) be given by f ( z ) = 1 + 9 e √ z + 9 e √ z + e √ z − e √ z + 9 e √ z + e √ z − e ( √ √ z − e (2 √ √ z − e ( √ √ z .The group Ξ f has rank r = 2, the numbers √ √ γ denotes the associated isomorphism of Ξ f onto Z one obtains - P ( ζ , ζ ) = 1 + 9 ζ + 9 ζ + ζ − ζ + 9 ζ + ζ − ζ ζ − ζ ζ − ζ ζ , - < υ ( f, γ ) = 2 − π (cid:0) √ <
142 , - f = 10 . Γ P Figure 3: 5 = ρ ( f ) (cid:54) f = 10.19 xample 4.4 Let f ∈ Exp( C ) be given by f ( z ) = 2 + 9 e √ z + 9 e √ z + e √ z + e √ z + 180 e ( √ √ z + 9 e (2 √ √ z + 9 e ( √ √ z + e √ √ z . Thegroup Ξ f has rank r = 2, the numbers √ √ γ denotes the associated isomorphism of Ξ f on Z one gets - P ( ζ , ζ ) = 2 + 9 ζ + 9 ζ + ζ + ζ + 180 ζ ζ + 9 ζ ζ + 9 ζ ζ + ζ ζ , - < υ ( f, γ ) = 2 − π (cid:0) √ <
70 , - f = 9 . Γ P Figure 4: 3 = ρ ( f ) (cid:54) f = 9. Example 4.5
Let f ∈ Exp( C ) be given by f ( z , z ) = 2 + 7 e z + 9 e z + e z +4 z + 18 e z +2 z . The group Ξ f has rank r = 2, the elements (2 , ,
2) generate it freely and if γ denotes the associated isomorphism ofΞ f on Z one gets - P ( ζ , ζ ) = 2 + 7 ζ + 9 ζ − ζ + ζ + 18 ζ , - < υ ( f, γ ) = 2 − π (cid:0) √ <
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